我从 http://www.movable-type.co.uk 实现了轴承"公式/scripts/latlong.html.但这似乎非常不准确 - 我怀疑我的实施中有一些错误.你能帮我找到它吗?我的代码如下:
I implemented the "bearing" formula from http://www.movable-type.co.uk/scripts/latlong.html. But it seems highly inaccurate - I suspect some mistakes in my implementation. Could you help me with finding it? My code is below:
protected static double bearing(double lat1, double lon1, double lat2, double lon2){
double longDiff= lon2-lon1;
double y = Math.sin(longDiff)*Math.cos(lat2);
double x = Math.cos(lat1)*Math.sin(lat2)-Math.sin(lat1)*Math.cos(lat2)*Math.cos(longDiff);
return Math.toDegrees((Math.atan2(y, x))+360)%360;
}
你只是把括号 ()
放错地方了.
You just have your parentheses ()
in the wrong place.
您正在为弧度值添加度数,这不起作用.toDegrees()
将为您完成从弧度到度数的转换,然后一旦您有度数的值,您就可以进行标准化.
You are adding degrees to a value in radians, which won't work. toDegrees()
will do the conversion from radians to degrees for you, then you do the normalisation once you have a value in degrees.
你有:
Math.toDegrees( (Math.atan2(y, x))+360 ) % 360;
但你需要:
( Math.toDegrees(Math.atan2(y, x)) + 360 ) % 360;
还请记住,Math.sin()
、Math.cos()
和所有其他三角函数的所有输入都必须以弧度表示.如果您的输入是度数,您需要先使用 Math.toRadians()
进行转换.
Remember also that all inputs to Math.sin()
, Math.cos()
and all the other trigonometric functions must be in radians. If your inputs are degrees you'll need to convert them using Math.toRadians()
first.
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